2019 - JAMB Mathematics Past Questions and Answers - page 1

\(\frac{mn}{a^2} - \frac{pq}{b^2} = 1\)

\(\frac{mn}{a^2} - 1 = \frac{pq}{b^2}\)

\(\frac{mn - a^2}{a^2} = \frac{pq}{b^2}\)

\(pq = \frac{b^2 (mn - a^2)}{a^2}\)

\(q = \frac{b^2(mn - a^2)}{a^2 p}\)

\(\tan 78 = \frac{h}{60}\)

\(h = 60 \tan 78\)

\(h = 60 \times 4.705 = 282.27m\)

\(\approxeq\) 282m 

\(m \otimes n = mn + m - n\)

3 \(\otimes\) (2 \(\otimes\) 4)

2 \(\otimes\) 4 = 2(4) + 2 - 4 = 6

3 \(otimes\) 6 = 3(6) + 3 - 6  = 15

Number of pupils = 4 + 13 + 30 + 44 + 9 = 100

Those of 8 - year olds = 13

On the pie chart, the angle represented by the 8-year olds = \(\frac{13}{100} \times 360°\)

= 46.8°

n(Total) = 50

n(Physics) = 40

n(Biology) = 30

Let x = n(Physics and Biology)

Hence, 

n(Physics only) = 40 -x

n(Biology only) = 30 - x

40 - x + 30 - x + x = 50

70 - x = 50

x = 20

\(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}}\)

= \((\frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}})(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} + \sqrt{3}})\)

= \(\frac{2 + \sqrt{6} + \sqrt{6} + 3}{2 - \sqrt{6} + \sqrt{6} - 3}\)

= \(\frac{5 + 2\sqrt{6}}{-1}\)

= \(- 5 - 2\sqrt{6}\)

Length of chord, l = \(2r \sin (\frac{\theta}{2})\)

= \(2(3) \sin (\frac{68}{2})\)

= \(6 \sin 34\)

= \(6 \times 0.559\)

= 3.354 cm

l = \(\approxeq\)  3.4 cm

\(\sin \theta = \cos (90 - \theta)\)

\(\sin \theta = \cos (90 - 58)\)

= \(\cos 32\)

\(\frac{\frac{2}{3} \div \frac{4}{5}}{\frac{1}{4} + \frac{3}{5} - \frac{1}{3}}\)

\(\frac{2}{3} \div \frac{4}{5} = \frac{2}{3} \times \frac{5}{4}\)

= \(\frac{5}{6}\)

\(\frac{1}{4} + \frac{3}{5} - \frac{1}{3} = \frac{15 + 36 - 20}{60}\)

= \(\frac{31}{60}\)

\(\therefore \frac{\frac{2}{3} \div \frac{4}{5}}{\frac{1}{4} + \frac{3}{5} - \frac{1}{3}} = \frac{5}{6} \div \frac{31}{60}\)

= \(\frac{5}{6} \times \frac{60}{31}\)

= \(\frac{50}{31}\)